C3H7NO3A -Amino acidC4H10NO3ClBC4H9NO2Cl2CC4H9NO2Cl2CC3H6NO2ClDAlanineCompound D is -

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Amino Acids

C3H7NO3A α-Am...

Question

$(A) α - Amino acid (C_{3} H_{7} N O_{3}) M e O H - --- \to H C l (B) (C_{4} H_{10} N O_{3} C l) P C l_{5} - -- \to (C) (C_{4} H_{9} N O_{2} C l_{2})$

$(C) (C_{4} H_{9} N O_{2} C l_{2}) H_{3} O^{+} - -- \to (D) (C_{3} H_{6} N O_{2} C l) N a - H g - ---- \to H C l Alanine$

Compound (D) is :

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Both (b) and (c)

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Solution

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(A) α - Amino acid (C_{3} H_{7} N O_{3}) M e O H - --- \to H C l (B) (C_{4} H_{10} N O_{3} C l) P C l_{5} - -- \to (C) (C_{4} H_{9} N O_{2} C l_{2})

(C) (C_{4} H_{9} N O_{2} C l_{2}) H_{3} O^{+} - -- \to (D) (C_{3} H_{6} N O_{2} C l) N a - H g - ---- \to H C l Alanine

The enantiomer of $α - D - (+) -$ glucose is ?

α - D (+)

β - D (+)

{[α]}_{D}

α

α - D - g l u c o s e_{(s)} + H_{2} O ⟶ α - D - g l u c o s e_{(a q)}; Δ H_{d i s s o l u t i o n} = 10.84 k J

β - D - g l u c o s e_{(s)} + H_{2} O ⟶ β - D - g l u c o s e_{(a q)}; Δ H_{d i s s o l u t i o n} = 4.68 k J

α - D - g l u c o s e_{(a q)} ⟶ β - D - g l u c o s e_{(a q)}; Δ H_{m u t a r o t a t i o n} = - 1.16 k J

Δ H^{\circ}

α - D - g l u c o s e ⟶ β - D - g l u c o s e

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